Calculate the enthalpy of breathing at isobaric conditions for a man that inhale

Question

Calculate the enthalpy of breathing at isobaric conditions for a man that inhales hot air (110°F ? 316K) and exhales air at (98.5°F ? 310 K) incoporating the provided data set of Specific Heat at constant pressure below. In addition to estimating the amount of heat absorbed in one day by the air breathed by the man

Average volume of air in 1 breath = 230 cm3

Density of Air= 0.0012928 g/cm3

*Assume 10 breaths/min.*

Resulting Answer must be in kJ/Day

Data Set Below

Temperature in Kelvin 114.5 102.45 103.44 105.33 7 95.33 96 33 38.23 10 92.32 93 32 95 23 8141 82.45 84.43 50 78 14 79.21 81-23 60 642 51 3.56 70 74.93 7604 7812 80 73 64 24.76 75.87 73.62 75.75 100 7141 7257 7473 69.54 140 6792 69.14 71.4 160 6649 67.74 70.05 250 67 62.99 65.44 63.54 61.99 59.45 400 5675 5811 60 66 600 52.781 5412 56.65

Explanation / Answer

mass flow of air m= 0.0012928 g/cm3*230cm^3 =297.344 mg per 1 breath

=> m=297.344*10^-3 g / 1 breath*10 breaths/min =2.97344 kg/min =2.97344 *1,440 kg/day =4281.7536 kg/day

Heat absorbed Q= m*Cp*(T2-T1) =4281.7536*114.5*(316-310) =2941564.7232 ( units as of Cp, which is not visible in image uploaded in question)

Assuming Cp is in J/kg.K => Q= 2941564.7232 J/day =2941.56kJ/day

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